Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $q = \dfrac{-4n^2 + 4n}{n^3 - 2n^2 - 8n} \div \dfrac{-3n + 27}{-5n^2 + 65n - 180} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4n^2 + 4n}{n^3 - 2n^2 - 8n} \times \dfrac{-5n^2 + 65n - 180}{-3n + 27} $ First factor out any common factors. $q = \dfrac{-4n(n - 1)}{n(n^2 - 2n - 8)} \times \dfrac{-5(n^2 - 13n + 36)}{-3(n - 9)} $ Then factor the quadratic expressions. $q = \dfrac {-4n(n - 1)} {n(n - 4)(n + 2)} \times \dfrac {-5(n - 4)(n - 9)} {-3(n - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-4n(n - 1) \times -5(n - 4)(n - 9) } { n(n - 4)(n + 2) \times -3(n - 9)} $ $q = \dfrac {20n(n - 4)(n - 9)(n - 1)} {-3n(n - 4)(n + 2)(n - 9)} $ Notice that $(n - 4)$ and $(n - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {20n\cancel{(n - 4)}(n - 9)(n - 1)} {-3n\cancel{(n - 4)}(n + 2)(n - 9)} $ We are dividing by $n - 4$ , so $n - 4 \neq 0$ Therefore, $n \neq 4$ $q = \dfrac {20n\cancel{(n - 4)}\cancel{(n - 9)}(n - 1)} {-3n\cancel{(n - 4)}(n + 2)\cancel{(n - 9)}} $ We are dividing by $n - 9$ , so $n - 9 \neq 0$ Therefore, $n \neq 9$ $q = \dfrac {20n(n - 1)} {-3n(n + 2)} $ $ q = \dfrac{-20(n - 1)}{3(n + 2)}; n \neq 4; n \neq 9 $